# How do I read this distribution function: $\min(X,Y)$?

I'm confused on what the $\min$ means. For example if I need to find the distribution function of $\min(X,Y)$ what am I looking for exactly? Am I looking for the distribution of the minimum value of either $X$ or $Y$. As an example lets just say that $X$ can have the values $\{1,2,3,5,8,13\}$ and $Y$ can have the values $\{3,4,6,7,8,12,24\}$. If were finding the $\min(X,Y)$, how do I do that? I think this is a legitimate example, if not you can fill in the blanks or use whatever you need to. I just dont understand when I see a problem ask for the distribution of $\min(X,Y)$.

I'm aware that theres also $\max(X,Y)$ as well but I'm just choosing $\min$ for the sake of this post. I'm assuming the same procedure goes for $\max$ as $\min$.

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$min(X,Y)$ is defined by $min(X,Y)(\omega)=min(X(\omega),Y(\omega))$ for every $\omega\in\Omega$. Does that help? –  anonymous Nov 10 '12 at 18:56

Say you toss a pair of fair three-sided dice. The following outcomes are equally probable: $$\begin{array}{ccc} 1,1 & 1,2 & 1,3 \\ 2,1 & 2,2 & 2,3 \\ 3,1 & 3,2 & 3,3 \end{array}$$ Call the first component in each pair $X$ and the second $Y$. Then $\min\{X,Y\}$ takes these values: $$\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array}$$ So the probability distribution is given by: \begin{align} \Pr(\min\{X,Y\} = 1) & = 5/9 \\[6pt] \Pr(\min\{X,Y\} = 2) & = 3/9 \\[6pt] \Pr(\min\{X,Y\} = 3) & = 1/9 \end{align}

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I like to look at the min this way:

Say the random variable $Z = \min(X,Y)$

$P\{Z \le t\} = P\{ (X \le t) \cup (Y \le t)\}$

Then remember that $P\{Z \le t\}$ is definition of the distribution $F_z(t)$

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